bisection method using matlab

vertical velocity of a motorcycle due to a road bump is given by:

V(t)=X*exp(-zeta*Wn*t)*(-zeta*Wn*sin(Wd*t)+Wd*cos(Wd*t))

Where; X=maximum displacement of the motor cycle , 0.4550m zeta=damping constant , 0.4037 Wn=undamped natural frequency of the system , 3.4338rad/s Wd= Wn*sqrt(1-zeta^2)=damped natural frequency of the system t=time

Determine the time, t, at which the velocity of the motorcycle attains a value of 1m/s.

I know how to find it by using microsoft excel. However, I would like to try it in matlab. Can anyone help me?

Thx

[bym]

you usually have to show some attempt at using matlab, and ask a specific question to get a response

Answer 1

[Niranjan Sundararajan]

Hey there,

What you essentially want to compute is the inverse of the function V(t). To do so in MATLAB, you can use the fzero function. Documentation link - https://in.mathworks.com/help/matlab/ref/fzero.html

Now, you want to find V(inverse) of 1 m/s. To do so, follow the following code snippet. P.S. - I have used the formula you provided for velocity.

format long;
V=1; %m/s
f_inv_val_of_t = fzero(@(t) f(t) - V, 0) %seconds

f_inv_val_of_t = 
   0.103244252227968

verification_output = f(f_inv_val_of_t) % = 1 m/s only

verification_output = 
   1.000000000000000

function V = f(t)
    %time in seconds
    X = 0.4550; %m
    zeta = 0.4037;
    Wn = 3.4338; %rad/s
    Wd = Wn*sqrt(1-zeta^2); %rad/s
    V=X*exp(-zeta*Wn*t)*(-zeta*Wn*sin(Wd*t)+Wd*cos(Wd*t)); %m/s
end